The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((94, 97)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-94)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(90,94)
2
(38,55)
3
(99,109)
4
(64,57)
5
(114,76)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-94)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(90-94)^2+(94-97)^2}~=~5\]
\[\sqrt{(38-94)^2+(55-97)^2}~=~70\]
\[\sqrt{(99-94)^2+(109-97)^2}~=~13\]
\[\sqrt{(64-94)^2+(57-97)^2}~=~50\]
\[\sqrt{(114-94)^2+(76-97)^2}~=~29\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((102, 105)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-105)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(126,112)
2
(142,114)
3
(22,45)
4
(120,81)
5
(105,109)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-105)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(126-102)^2+(112-105)^2}~=~25\]
\[\sqrt{(142-102)^2+(114-105)^2}~=~41\]
\[\sqrt{(22-102)^2+(45-105)^2}~=~100\]
\[\sqrt{(120-102)^2+(81-105)^2}~=~30\]
\[\sqrt{(105-102)^2+(109-105)^2}~=~5\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((103, 98)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(43,123)
2
(43,178)
3
(113,122)
4
(52,30)
5
(58,74)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(43-103)^2+(123-98)^2}~=~65\]
\[\sqrt{(43-103)^2+(178-98)^2}~=~100\]
\[\sqrt{(113-103)^2+(122-98)^2}~=~26\]
\[\sqrt{(52-103)^2+(30-98)^2}~=~85\]
\[\sqrt{(58-103)^2+(74-98)^2}~=~51\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((91, 101)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-91)^2+(y_2-101)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(61,173)
2
(131,92)
3
(154,161)
4
(103,66)
5
(109,181)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-91)^2+(y_2-101)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(61-91)^2+(173-101)^2}~=~78\]
\[\sqrt{(131-91)^2+(92-101)^2}~=~41\]
\[\sqrt{(154-91)^2+(161-101)^2}~=~87\]
\[\sqrt{(103-91)^2+(66-101)^2}~=~37\]
\[\sqrt{(109-91)^2+(181-101)^2}~=~82\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((104, 107)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-104)^2+(y_2-107)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(92,116)
2
(83,35)
3
(71,163)
4
(124,59)
5
(8,135)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-104)^2+(y_2-107)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(92-104)^2+(116-107)^2}~=~15\]
\[\sqrt{(83-104)^2+(35-107)^2}~=~75\]
\[\sqrt{(71-104)^2+(163-107)^2}~=~65\]
\[\sqrt{(124-104)^2+(59-107)^2}~=~52\]
\[\sqrt{(8-104)^2+(135-107)^2}~=~100\]
Question
Using Desmos, determine which of the given points are within a distance of 11 units from point (3, 5).
Graph the inequality \((x-3)^2+(y-5)^2~\le~11^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(12,11)
(11,14)
(7,16)
(-1,14)
(-5,12)
(-8,2)
(-7,1)
(-2,-5)
(8,-5)
(15,2)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (12,11) is in
point (11,14) is NOT in
point (7,16) is NOT in
point (-1,14) is in
point (-5,12) is in
point (-8,2) is NOT in
point (-7,1) is in
point (-2,-5) is NOT in
point (8,-5) is NOT in
point (15,2) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 11 units from point (6, 4).
Graph the inequality \((x-6)^2+(y-4)^2~\le~11^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(17,9)
(9,13)
(4,15)
(-1,11)
(-3,10)
(-3,0)
(-3,-5)
(5,-7)
(12,-6)
(18,2)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (17,9) is NOT in
point (9,13) is in
point (4,15) is NOT in
point (-1,11) is in
point (-3,10) is in
point (-3,0) is in
point (-3,-5) is NOT in
point (5,-7) is NOT in
point (12,-6) is NOT in
point (18,2) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 8 units from point (-6, 1).
Graph the inequality \((x+6)^2+(y-1)^2~\le~8^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(3,3)
(-2,7)
(-6,8)
(-9,7)
(-12,4)
(-14,-2)
(-9,-6)
(-3,-8)
(2,-2)
(2,-1)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (3,3) is NOT in
point (-2,7) is in
point (-6,8) is in
point (-9,7) is in
point (-12,4) is in
point (-14,-2) is NOT in
point (-9,-6) is in
point (-3,-8) is NOT in
point (2,-2) is NOT in
point (2,-1) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 15 units from point (5, -1).
Graph the inequality \((x-5)^2+(y+1)^2~\le~15^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(18,9)
(16,9)
(4,14)
(-6,10)
(-10,-2)
(-4,-13)
(3,-17)
(9,-14)
(20,-7)
(20,-3)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
point (18,9) is NOT in
point (16,9) is in
point (4,14) is NOT in
point (-6,10) is NOT in
point (-10,-2) is NOT in
point (-4,-13) is in
point (3,-17) is NOT in
point (9,-14) is in
point (20,-7) is NOT in
point (20,-3) is NOT in
Question
Using Desmos, determine which of the given points are within a distance of 8 units from point (-2, -3).
Graph the inequality \((x+2)^2+(y+3)^2~\le~8^2\) as the first item in Desmos.
Enter each point as a Cartesian coordinate pair. Actually, you should be able to copy/paste all the points as a single item (using commas to separate the points):
Determine which points are in the shaded area. If a point is exactly on the boundary, count that point as within the distance. You might need to zoom in on points to be sure.
(6,-3)
(4,5)
(1,6)
(-6,4)
(-10,2)
(-11,-4)
(-9,-5)
(-6,-9)
(2,-10)
(5,-8)
Solution
Follow the directions given in the prompt. I’ve plotted the inequality and the points below, using two different marks to indicate whether the point in inside the boundary or outside the boundary. Notice the points should be listed (mostly) in a counter-clockwise order, starting from 3 o’clock (see standard position of an angle).
As the first item, use the standard form of a circle to visualize the first criterion.
\[(x-73)^2+(y-31)^2~\le~25^2\]
As the second item, use the standard form of a circle to visualize the second criterion.
\[(x-39)^2+(y-54)^2~\le~24^2\]
As the third item, copy/paste the list of points. Click the “Label” checkbox.
Read which points are in the overlapping region.
point (61,40) is NOT in
point (55,50) is NOT in
point (48,37) is NOT in
point (57,45) is in
point (53,49) is NOT in
point (63,44) is NOT in
point (62,34) is NOT in
point (60,46) is in
point (58,48) is in
point (52,52) is NOT in
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (94, 99).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (95, 100).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (104, 100).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (95, 104).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle in an \(xy\)-plane.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (106, 108).
Question
Consider the triangle with vertices at the following coordinates:
(9, 1)
(2, 11)
(15, 14)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taking differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{7^2+10^2}~=~ 12.2065556\]
\[\sqrt{13^2+3^2}~=~ 13.3416641\]
\[\sqrt{6^2+13^2}~=~ 14.3178211\]
To find the perimeter, add up the three lengths.
\[P~=~12.2065556+13.3416641+14.3178211\]
\[P~=~39.8660407\]
Question
Consider the triangle with vertices at the following coordinates:
(12, 4)
(14, 9)
(3, 11)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taking differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{2^2+5^2}~=~ 5.3851648\]
\[\sqrt{11^2+2^2}~=~ 11.1803399\]
\[\sqrt{9^2+7^2}~=~ 11.4017543\]
To find the perimeter, add up the three lengths.
\[P~=~5.3851648+11.1803399+11.4017543\]
\[P~=~27.9672589\]
Question
Consider the triangle with vertices at the following coordinates:
(16, 19)
(3, 9)
(11, 1)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taking differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{13^2+10^2}~=~ 16.4012195\]
\[\sqrt{8^2+8^2}~=~ 11.3137085\]
\[\sqrt{5^2+18^2}~=~ 18.6815417\]
To find the perimeter, add up the three lengths.
\[P~=~16.4012195+11.3137085+18.6815417\]
\[P~=~46.3964697\]
Question
Consider the triangle with vertices at the following coordinates:
(13, 17)
(8, 10)
(15, 4)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taking differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{5^2+7^2}~=~ 8.6023253\]
\[\sqrt{7^2+6^2}~=~ 9.2195445\]
\[\sqrt{2^2+13^2}~=~ 13.1529464\]
To find the perimeter, add up the three lengths.
\[P~=~8.6023253+9.2195445+13.1529464\]
\[P~=~30.9748162\]
Question
Consider the triangle with vertices at the following coordinates:
(6, 1)
(19, 4)
(14, 11)
Calculate the perimeter. Please be accurate to within 0.01 units.
Solution
We want to use the Pythagorean theorem, which tells us the hypotenuse’s length, \(c\), can be calculated from the two legs’ lengths (\(a\) and \(b\)) by using \(c=\sqrt{a^2+b^2}\). It is easy to draw right triangles by using vertical and horizontal legs. The leg lengths can be found by taking differences of corresponding coordinates or by counting on the graph. See the diagram below.
Calculate the hypotenuses’ lengths.
\[\sqrt{13^2+3^2}~=~ 13.3416641\]
\[\sqrt{5^2+7^2}~=~ 8.6023253\]
\[\sqrt{8^2+10^2}~=~ 12.8062485\]
To find the perimeter, add up the three lengths.
\[P~=~13.3416641+8.6023253+12.8062485\]
\[P~=~34.7502378\]
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
13
-15
-19
-63
16
-9
-13
85
-19
-24
26
44
23
19
-55
-20
-51
-17
-43
-58
44
31
-50
1
63
10
-17
-37
-22
55
49
36
-1
-59
-38
21
-30
11
-14
-34
-35
-6
64
-1
2
-17
17
20
61
-35
-70
-11
-46
27
11
-37
27
21
35
-55
-18
33
-69
-20
-21
15
-7
20
50
-25
-29
-25
How many of these arrows landed in the ring worth 5 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 5 points has an inner radius of 50 mm and an outer radius of 60 mm. So, we want to know how many arrows have distances (from origin) between 50 mm and 60 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 5 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 13 in cell A2 and -15 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 50 mm away, use =IF(C2>50,1,0) in cell D2.
To check if the arrow is less than 60 mm away, use =IF(C2<60,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 50 mm
is less than 60 mm
is in ring
how many in
13
-15
19.84943
0
1
0
9
-19
-63
65.80274
1
0
0
16
-9
18.35756
0
1
0
-13
85
85.98837
1
0
0
-19
-24
30.61046
0
1
0
26
44
51.10773
1
1
1
23
19
29.83287
0
1
0
-55
-20
58.52350
1
1
1
-51
-17
53.75872
1
1
1
-43
-58
72.20111
1
0
0
44
31
53.82379
1
1
1
-50
1
50.01000
1
1
1
63
10
63.78871
1
0
0
-17
-37
40.71855
0
1
0
-22
55
59.23681
1
1
1
49
36
60.80296
1
0
0
-1
-59
59.00847
1
1
1
-38
21
43.41659
0
1
0
-30
11
31.95309
0
1
0
-14
-34
36.76955
0
1
0
-35
-6
35.51056
0
1
0
64
-1
64.00781
1
0
0
2
-17
17.11724
0
1
0
17
20
26.24881
0
1
0
61
-35
70.32780
1
0
0
-70
-11
70.85901
1
0
0
-46
27
53.33854
1
1
1
11
-37
38.60052
0
1
0
27
21
34.20526
0
1
0
35
-55
65.19202
1
0
0
-18
33
37.58989
0
1
0
-69
-20
71.84010
1
0
0
-21
15
25.80698
0
1
0
-7
20
21.18962
0
1
0
50
-25
55.90170
1
1
1
-29
-25
38.28838
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 5 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
28
5
-27
26
-54
-19
-15
8
-51
36
29
1
12
35
14
-66
13
59
37
5
12
42
61
34
-12
6
20
-47
-62
11
4
3
68
34
-76
12
9
-66
65
-48
25
-11
24
-62
19
49
19
-10
10
-44
29
40
22
5
-33
-24
-26
-27
-40
17
-13
-12
39
33
26
-57
-55
-36
12
81
17
-17
How many of these arrows landed in the ring worth 4 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 4 points has an inner radius of 60 mm and an outer radius of 70 mm. So, we want to know how many arrows have distances (from origin) between 60 mm and 70 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 4 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 28 in cell A2 and 5 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 60 mm away, use =IF(C2>60,1,0) in cell D2.
To check if the arrow is less than 70 mm away, use =IF(C2<70,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 60 mm
is less than 70 mm
is in ring
how many in
28
5
28.44293
0
1
0
9
-27
26
37.48333
0
1
0
-54
-19
57.24509
0
1
0
-15
8
17.00000
0
1
0
-51
36
62.42596
1
1
1
29
1
29.01724
0
1
0
12
35
37.00000
0
1
0
14
-66
67.46851
1
1
1
13
59
60.41523
1
1
1
37
5
37.33631
0
1
0
12
42
43.68066
0
1
0
61
34
69.83552
1
1
1
-12
6
13.41641
0
1
0
20
-47
51.07837
0
1
0
-62
11
62.96825
1
1
1
4
3
5.00000
0
1
0
68
34
76.02631
1
0
0
-76
12
76.94154
1
0
0
9
-66
66.61081
1
1
1
65
-48
80.80223
1
0
0
25
-11
27.31300
0
1
0
24
-62
66.48308
1
1
1
19
49
52.55473
0
1
0
19
-10
21.47091
0
1
0
10
-44
45.12206
0
1
0
29
40
49.40648
0
1
0
22
5
22.56103
0
1
0
-33
-24
40.80441
0
1
0
-26
-27
37.48333
0
1
0
-40
17
43.46263
0
1
0
-13
-12
17.69181
0
1
0
39
33
51.08816
0
1
0
26
-57
62.64982
1
1
1
-55
-36
65.73431
1
1
1
12
81
81.88406
1
0
0
17
-17
24.04163
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 4 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
6
-4
-66
42
10
35
20
90
49
61
38
-26
-71
27
44
-13
67
-7
49
-19
5
15
66
39
0
-12
61
6
-3
-20
34
69
38
-4
-11
5
-7
26
-40
-27
0
-54
-42
-75
-38
-35
-45
2
11
10
56
-19
-51
-3
59
10
-23
20
-12
25
21
68
5
-52
3
3
20
-20
17
-27
11
7
How many of these arrows landed in the ring worth 5 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 5 points has an inner radius of 50 mm and an outer radius of 60 mm. So, we want to know how many arrows have distances (from origin) between 50 mm and 60 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 5 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 6 in cell A2 and -4 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 50 mm away, use =IF(C2>50,1,0) in cell D2.
To check if the arrow is less than 60 mm away, use =IF(C2<60,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 50 mm
is less than 60 mm
is in ring
how many in
6
-4
7.211103
0
1
0
7
-66
42
78.230429
1
0
0
10
35
36.400549
0
1
0
20
90
92.195445
1
0
0
49
61
78.243211
1
0
0
38
-26
46.043458
0
1
0
-71
27
75.960516
1
0
0
44
-13
45.880279
0
1
0
67
-7
67.364679
1
0
0
49
-19
52.554733
1
1
1
5
15
15.811388
0
1
0
66
39
76.661594
1
0
0
0
-12
12.000000
0
1
0
61
6
61.294372
1
0
0
-3
-20
20.223748
0
1
0
34
69
76.922038
1
0
0
38
-4
38.209946
0
1
0
-11
5
12.083046
0
1
0
-7
26
26.925824
0
1
0
-40
-27
48.259714
0
1
0
0
-54
54.000000
1
1
1
-42
-75
85.959293
1
0
0
-38
-35
51.662365
1
1
1
-45
2
45.044423
0
1
0
11
10
14.866069
0
1
0
56
-19
59.135438
1
1
1
-51
-3
51.088159
1
1
1
59
10
59.841457
1
1
1
-23
20
30.479501
0
1
0
-12
25
27.730849
0
1
0
21
68
71.168813
1
0
0
5
-52
52.239832
1
1
1
3
3
4.242641
0
1
0
20
-20
28.284271
0
1
0
17
-27
31.906112
0
1
0
11
7
13.038405
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 5 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
46
31
67
57
63
-30
-3
-26
-41
-55
32
-32
-34
18
22
-19
-8
-5
14
47
-34
64
-10
-43
46
-41
-7
-71
24
-52
24
40
-23
19
39
17
-6
2
0
-51
45
-32
98
-10
-22
-25
17
21
-38
4
3
48
24
7
-2
-30
-63
-54
16
-17
-11
43
21
-27
49
-46
-23
-21
-60
-15
-25
13
How many of these arrows landed in the ring worth 6 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 6 points has an inner radius of 40 mm and an outer radius of 50 mm. So, we want to know how many arrows have distances (from origin) between 40 mm and 50 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Copy and paste the given table into the spreadsheet.
Use the distance formula to calculate each arrow’s distance from origin.
Use IF statements to check if distance matches the ring.
Count up how many arrows are in the ring worth 6 points.
I’ll assume you have pasted the first arrow’s position in the second row, so 46 in cell A2 and 31 in cell B2.
You could use =SQRT(A2^2+B2^2) in cell C2 to find the distance of the first arrow from the origin.
To check if the arrow is more than 40 mm away, use =IF(C2>40,1,0) in cell D2.
To check if the arrow is less than 50 mm away, use =IF(C2<50,1,0) in cell E2.
To check if the arrow is in the ring, use =D2*E2 in cell F2.
You should then drag all these formulas down to apply them to every arrow.
To count how many arrows are in the ring, I’d use =sum(F2:F25) in cell G2.
The results should looks something like:
x (mm)
y (mm)
distance
is more than 40 mm
is less than 50 mm
is in ring
how many in
46
31
55.470713
1
0
0
7
67
57
87.965902
1
0
0
63
-30
69.778220
1
0
0
-3
-26
26.172505
0
1
0
-41
-55
68.600291
1
0
0
32
-32
45.254834
1
1
1
-34
18
38.470768
0
1
0
22
-19
29.068884
0
1
0
-8
-5
9.433981
0
1
0
14
47
49.040799
1
1
1
-34
64
72.470684
1
0
0
-10
-43
44.147480
1
1
1
46
-41
61.619802
1
0
0
-7
-71
71.344236
1
0
0
24
-52
57.271284
1
0
0
24
40
46.647615
1
1
1
-23
19
29.832868
0
1
0
39
17
42.544095
1
1
1
-6
2
6.324555
0
1
0
0
-51
51.000000
1
0
0
45
-32
55.217751
1
0
0
98
-10
98.508883
1
0
0
-22
-25
33.301652
0
1
0
17
21
27.018512
0
1
0
-38
4
38.209946
0
1
0
3
48
48.093659
1
1
1
24
7
25.000000
0
1
0
-2
-30
30.066593
0
1
0
-63
-54
82.975900
1
0
0
16
-17
23.345235
0
1
0
-11
43
44.384682
1
1
1
21
-27
34.205263
0
1
0
49
-46
67.208630
1
0
0
-23
-21
31.144823
0
1
0
-60
-15
61.846584
1
0
0
-25
13
28.178006
0
1
0
We can also visualize the arrow positions, and indicate whether the arrow is in the ring worth 6 points using a filled dot.
Question
An archery target is 200 millimeters wide. It is composed of concentric rings. The center ring is worth 10 points, the next ring is worth 9, and so on as shown in the diagram. Notice the radii of the rings are all multiples of 10 millimeters.
By setting up a Cartesian coordinate system, with units of millimeters, and the origin at the bullseye, we can indicate any point on the target with a Cartesian-coordinate ordered pair.
An archer shoots 36 arrows; the positions are indicated in the table below.
x (mm)
y (mm)
6
-14
27
3
-40
21
-31
-2
5
18
-18
17
-43
52
-11
65
-61
45
6
15
50
18
-21
-28
0
74
-11
-3
14
-1
-55
50
-27
-39
-39
50
-16
8
-14
17
3
21
-17
-82
31
-51
9
94
6
41
-30
86
98
19
-29
-24
-16
15
-79
31
-58
62
-45
62
31
-5
40
13
1
9
27
-88
How many of these arrows landed in the ring worth 3 points?
(Assume the arrows are infinitely skinny. None landed exactly on a boundary.)
Solution
Each arrow has a distance from the origin. Notice the ring worth 3 points has an inner radius of 70 mm and an outer radius of 80 mm. So, we want to know how many arrows have distances (from origin) between 70 mm and 80 mm.
Notice, because the origin is at (0,0), the distance formula simplifies.
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
Please be accurate within \(\pm\) 0.01 of the exact answer.
Solution
Let’s use Pythagorean Theorem. This is possible because all angles are 90 degrees. To do this, we can first find the length of a diagonal across the bottom. Let’s call it \(B\).
\[B~=~\sqrt{12^2+9^2}~=~\sqrt{225} ~\approx~15\]
Now we can form another right triangle, using \(B\), the height, and the desired length, which we can call \(D\).
where \(\Delta x=x_2-x_1\) and \(\Delta y=y_2-y_1\) and \(\Delta z=z_2-z_1\).
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+6x+y^2+10y~=~-18\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+6x+y^2+10y~=~-18\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+6x+9+y^2+10y+25~=~-18+9+25\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+3)^2+(y+5)^2~=~16\]
Notice \(16\) is a perfect square integer.
\[(x+3)^2+(y+5)^2~=~4^2\]
Thus,
\[h=-3\]\[k=-5\]\[r=4\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-16x+y^2-8y~=~-71\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-16x+y^2-8y~=~-71\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-16x+64+y^2-8y+16~=~-71+64+16\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-8)^2+(y-4)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x-8)^2+(y-4)^2~=~3^2\]
Thus,
\[h=8\]\[k=4\]\[r=3\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2-12x+y^2-8y~=~-43\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2-12x+y^2-8y~=~-43\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2-12x+36+y^2-8y+16~=~-43+36+16\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x-6)^2+(y-4)^2~=~9\]
Notice \(9\) is a perfect square integer.
\[(x-6)^2+(y-4)^2~=~3^2\]
Thus,
\[h=6\]\[k=4\]\[r=3\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+12x+y^2-14y~=~-81\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+12x+y^2-14y~=~-81\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+12x+36+y^2-14y+49~=~-81+36+49\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+6)^2+(y-7)^2~=~4\]
Notice \(4\) is a perfect square integer.
\[(x+6)^2+(y-7)^2~=~2^2\]
Thus,
\[h=-6\]\[k=7\]\[r=2\]
Question
When plotting all \(x\)-\(y\) pairs that satisfy the given equation (make the equation true), the result is a circle.
\[x^2+4x+y^2-8y~=~16\]
The standard form of an equation of a circle shows the center \((h,k)\) and radius \(r\) as parameters.
\[(x-h)^2+(y-k)^2~=~r^2\]
In order to convert the given circle into standard form, you should complete the square (twice, once for \(x\)-containing terms and once for \(y\)-containing terms). After you convert the equation to standard form, indicate the values of the parameters.
\(h=\)
\(k=\)
\(r=\)
Solution
The given equation:
\[x^2+4x+y^2-8y~=~16\]
Determine the constants needed for perfect squares. Do this by halving the linear coefficient and squaring the result. In other words, \(z^2+bz+\left(\frac{b}{2}\right)^2\) is a perfect square for any value of \(b\).
\[x^2+4x+4+y^2-8y+16~=~16+4+16\]
Write each perfect square in factored form. Also, simplify the right side.
\[(x+2)^2+(y-4)^2~=~36\]
Notice \(36\) is a perfect square integer.
\[(x+2)^2+(y-4)^2~=~6^2\]
Thus,
\[h=-2\]\[k=4\]\[r=6\]
Question
Consider the two points:
Point \(A\) is at \((18,81)\)
Point \(B\) is at \((48,21)\)
A third collinear point, \(C\), is 40% of the way from point \(A\) to point \(B\), as shown below.
Find the coordinates of point \(C\):
(,)
Solution
There are many different ways to approach this problem.
Let’s start by finding the deltas: \(\Delta x\) and \(\Delta y\).
\[\Delta x ~=~ 48-18 ~=~ 30\]\[\Delta y ~=~ 21-81 ~=~ -60\]
The absolute values of these deltas are lengths of legs of a right triangle with vertical and horizontal legs.
We can multiply each absolute delta by the given percentage.
\[0.4 \cdot 30 ~=~ 12\]\[0.4 \cdot 60 ~=~ 24\]
We can draw a new right triangle, starting at \(A\), with these leg lengths. Point \(C\) should be on a vertex of this new triangle.
A taxi service charges a fixed base cost of $2.72 and then a mileage cost of $1.75 per mile. The following people wonder whether they can travel a specified distance with the amount of cash they are carrying.
Can Arnold go 18 miles with $28.61?
Can Betty go 5.5 miles with $9.49?
Can Chester go 10.3 miles with $25.93?
Can Denali go 22.4 miles with $33.68?
Can Edward go 26.4 miles with $63.40?
Solution
I’d recommend doing this with Desmos. Start by creating the equation describing the linear equation between distance (\(x\), miles) and total cost (\(y\), dollars).
\[y~=~2.72+1.75x\]
A passenger needs to have at least the total in cash to afford the ride. If the passenger has more than enough money, they can still afford the ride. So, if we redefine \(y\) to be any possible amount of pocket money, we can say \(y\) must be as much or more than the cost of the ride.
\[y~\ge~2.72+1.75x\]
So, type that inequality into Desmos. Then type in the following points:
(18,28.61)
(5.5,9.49)
(10.3,25.93)
(22.4,33.68)
(26.4,63.4)
For each point, if it falls in the feasible region, then it is a solution to the inequality.
No, Arnold can NOT afford to go 18 miles with $28.61 in pocket.
No, Betty can NOT afford to go 5.5 miles with $9.49 in pocket.
Yes, Chester can afford to go 10.3 miles with $25.93 in pocket.
No, Denali can NOT afford to go 22.4 miles with $33.68 in pocket.
Yes, Edward can afford to go 26.4 miles with $63.40 in pocket.
Question
A taxi service charges a fixed base cost of $6.00 and then a mileage cost of $2.39 per mile. The following people wonder whether they can travel a specified distance with the amount of cash they are carrying.
Can Arnold go 11.4 miles with $38.47?
Can Betty go 7.1 miles with $29.14?
Can Chester go 23.7 miles with $80.91?
Can Denali go 3.1 miles with $16.69?
Can Edward go 20.9 miles with $46.50?
Solution
I’d recommend doing this with Desmos. Start by creating the equation describing the linear equation between distance (\(x\), miles) and total cost (\(y\), dollars).
\[y~=~6.00+2.39x\]
A passenger needs to have at least the total in cash to afford the ride. If the passenger has more than enough money, they can still afford the ride. So, if we redefine \(y\) to be any possible amount of pocket money, we can say \(y\) must be as much or more than the cost of the ride.
\[y~\ge~6.00+2.39x\]
So, type that inequality into Desmos. Then type in the following points:
(11.4,38.47)
(7.1,29.14)
(23.7,80.91)
(3.1,16.69)
(20.9,46.5)
For each point, if it falls in the feasible region, then it is a solution to the inequality.
Yes, Arnold can afford to go 11.4 miles with $38.47 in pocket.
Yes, Betty can afford to go 7.1 miles with $29.14 in pocket.
Yes, Chester can afford to go 23.7 miles with $80.91 in pocket.
Yes, Denali can afford to go 3.1 miles with $16.69 in pocket.
No, Edward can NOT afford to go 20.9 miles with $46.50 in pocket.
Question
A taxi service charges a fixed base cost of $7.64 and then a mileage cost of $2.90 per mile. The following people wonder whether they can travel a specified distance with the amount of cash they are carrying.
Can Arnold go 16 miles with $65.69?
Can Betty go 6.5 miles with $32.13?
Can Chester go 26.5 miles with $69.97?
Can Denali go 12.5 miles with $35.79?
Can Edward go 20.3 miles with $82.79?
Solution
I’d recommend doing this with Desmos. Start by creating the equation describing the linear equation between distance (\(x\), miles) and total cost (\(y\), dollars).
\[y~=~7.64+2.90x\]
A passenger needs to have at least the total in cash to afford the ride. If the passenger has more than enough money, they can still afford the ride. So, if we redefine \(y\) to be any possible amount of pocket money, we can say \(y\) must be as much or more than the cost of the ride.
\[y~\ge~7.64+2.90x\]
So, type that inequality into Desmos. Then type in the following points:
(16,65.69)
(6.5,32.13)
(26.5,69.97)
(12.5,35.79)
(20.3,82.79)
For each point, if it falls in the feasible region, then it is a solution to the inequality.
Yes, Arnold can afford to go 16 miles with $65.69 in pocket.
Yes, Betty can afford to go 6.5 miles with $32.13 in pocket.
No, Chester can NOT afford to go 26.5 miles with $69.97 in pocket.
No, Denali can NOT afford to go 12.5 miles with $35.79 in pocket.
Yes, Edward can afford to go 20.3 miles with $82.79 in pocket.
Question
A taxi service charges a fixed base cost of $3.09 and then a mileage cost of $0.92 per mile. The following people wonder whether they can travel a specified distance with the amount of cash they are carrying.
Can Arnold go 4.2 miles with $8.01?
Can Betty go 22.6 miles with $27.71?
Can Chester go 26.8 miles with $22.13?
Can Denali go 15.1 miles with $11.38?
Can Edward go 7 miles with $10.85?
Solution
I’d recommend doing this with Desmos. Start by creating the equation describing the linear equation between distance (\(x\), miles) and total cost (\(y\), dollars).
\[y~=~3.09+0.92x\]
A passenger needs to have at least the total in cash to afford the ride. If the passenger has more than enough money, they can still afford the ride. So, if we redefine \(y\) to be any possible amount of pocket money, we can say \(y\) must be as much or more than the cost of the ride.
\[y~\ge~3.09+0.92x\]
So, type that inequality into Desmos. Then type in the following points:
(4.2,8.01)
(22.6,27.71)
(26.8,22.13)
(15.1,11.38)
(7,10.85)
For each point, if it falls in the feasible region, then it is a solution to the inequality.
Yes, Arnold can afford to go 4.2 miles with $8.01 in pocket.
Yes, Betty can afford to go 22.6 miles with $27.71 in pocket.
No, Chester can NOT afford to go 26.8 miles with $22.13 in pocket.
No, Denali can NOT afford to go 15.1 miles with $11.38 in pocket.
Yes, Edward can afford to go 7 miles with $10.85 in pocket.
Question
A taxi service charges a fixed base cost of $2.79 and then a mileage cost of $1.34 per mile. The following people wonder whether they can travel a specified distance with the amount of cash they are carrying.
Can Arnold go 14.1 miles with $25.09?
Can Betty go 19.7 miles with $20.34?
Can Chester go 5.1 miles with $11.00?
Can Denali go 22.9 miles with $27.97?
Can Edward go 27.1 miles with $27.75?
Solution
I’d recommend doing this with Desmos. Start by creating the equation describing the linear equation between distance (\(x\), miles) and total cost (\(y\), dollars).
\[y~=~2.79+1.34x\]
A passenger needs to have at least the total in cash to afford the ride. If the passenger has more than enough money, they can still afford the ride. So, if we redefine \(y\) to be any possible amount of pocket money, we can say \(y\) must be as much or more than the cost of the ride.
\[y~\ge~2.79+1.34x\]
So, type that inequality into Desmos. Then type in the following points:
(14.1,25.09)
(19.7,20.34)
(5.1,11)
(22.9,27.97)
(27.1,27.75)
For each point, if it falls in the feasible region, then it is a solution to the inequality.
Yes, Arnold can afford to go 14.1 miles with $25.09 in pocket.
No, Betty can NOT afford to go 19.7 miles with $20.34 in pocket.
Yes, Chester can afford to go 5.1 miles with $11.00 in pocket.
No, Denali can NOT afford to go 22.9 miles with $27.97 in pocket.
No, Edward can NOT afford to go 27.1 miles with $27.75 in pocket.
Question
A thief is stealing xots and yivs. Each xot has a mass of 15 kilograms and a volume of 8 liters. Each yiv has a mass of 5 kilograms and a volume of 12 liters. The thief can carry a maximum mass of 75 kilograms and a maximum volume of 96 liters. The profit from each xot is $6.54 and the profit from each yiv is $8.78.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
15
5
75
volume (L)
8
12
96
profit ($)
6.54
8.78
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here. But, we can add a third “dimension” by using a color ramp for profit. You can see that since our profit function is linear in \(x\) and linear in \(y\), the profit grows linearly as either xots or yivs are added. Another useful concept is an isoprofit line, which shows all the possible combinations of xots and yivs that would generate the same profit (see contour lines). Since profit is linear in both \(x\) and \(y\), we see the isoprofit lines are all parallel. Also, in this case, the isoprofit lines are not parallel to either constraint, so there will be only a single optimal choice.
Question
A thief is stealing xots and yivs. Each xot has a mass of 16 kilograms and a volume of 12 liters. Each yiv has a mass of 8 kilograms and a volume of 9 liters. The thief can carry a maximum mass of 128 kilograms and a maximum volume of 108 liters. The profit from each xot is $2.05 and the profit from each yiv is $6.66.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
16
8
128
volume (L)
12
9
108
profit ($)
2.05
6.66
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here. But, we can add a third “dimension” by using a color ramp for profit. You can see that since our profit function is linear in \(x\) and linear in \(y\), the profit grows linearly as either xots or yivs are added. Another useful concept is an isoprofit line, which shows all the possible combinations of xots and yivs that would generate the same profit (see contour lines). Since profit is linear in both \(x\) and \(y\), we see the isoprofit lines are all parallel. Also, in this case, the isoprofit lines are not parallel to either constraint, so there will be only a single optimal choice.
Question
A thief is stealing xots and yivs. Each xot has a mass of 8 kilograms and a volume of 9 liters. Each yiv has a mass of 20 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 160 kilograms and a maximum volume of 135 liters. The profit from each xot is $4.23 and the profit from each yiv is $8.94.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
8
20
160
volume (L)
9
15
135
profit ($)
4.23
8.94
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here. But, we can add a third “dimension” by using a color ramp for profit. You can see that since our profit function is linear in \(x\) and linear in \(y\), the profit grows linearly as either xots or yivs are added. Another useful concept is an isoprofit line, which shows all the possible combinations of xots and yivs that would generate the same profit (see contour lines). Since profit is linear in both \(x\) and \(y\), we see the isoprofit lines are all parallel. Also, in this case, the isoprofit lines are not parallel to either constraint, so there will be only a single optimal choice.
Question
A thief is stealing xots and yivs. Each xot has a mass of 18 kilograms and a volume of 20 liters. Each yiv has a mass of 6 kilograms and a volume of 4 liters. The thief can carry a maximum mass of 108 kilograms and a maximum volume of 80 liters. The profit from each xot is $1.29 and the profit from each yiv is $1.8.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
18
6
108
volume (L)
20
4
80
profit ($)
1.29
1.80
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here. But, we can add a third “dimension” by using a color ramp for profit. You can see that since our profit function is linear in \(x\) and linear in \(y\), the profit grows linearly as either xots or yivs are added. Another useful concept is an isoprofit line, which shows all the possible combinations of xots and yivs that would generate the same profit (see contour lines). Since profit is linear in both \(x\) and \(y\), we see the isoprofit lines are all parallel. Also, in this case, the isoprofit lines are not parallel to either constraint, so there will be only a single optimal choice.
Question
A thief is stealing xots and yivs. Each xot has a mass of 15 kilograms and a volume of 6 liters. Each yiv has a mass of 3 kilograms and a volume of 12 liters. The thief can carry a maximum mass of 45 kilograms and a maximum volume of 72 liters. The profit from each xot is $3.84 and the profit from each yiv is $5.17.
For your convenience, those numbers are organized in the table below.
attribute
xot
yiv
capacity
mass (kg)
15
3
45
volume (L)
6
12
72
profit ($)
3.84
5.17
\(\infty\)
What is the maximum profit the thief can produce? (In dollars.)
It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here. But, we can add a third “dimension” by using a color ramp for profit. You can see that since our profit function is linear in \(x\) and linear in \(y\), the profit grows linearly as either xots or yivs are added. Another useful concept is an isoprofit line, which shows all the possible combinations of xots and yivs that would generate the same profit (see contour lines). Since profit is linear in both \(x\) and \(y\), we see the isoprofit lines are all parallel. Also, in this case, the isoprofit lines are not parallel to either constraint, so there will be only a single optimal choice.
Question
Find the nearest distance from point \((-11,16)\) to line \(y=2x-22\).
For flavor, imagine on a map, a river can be approximated by the line \(y=2x-22\). You are currently at position \((-11,16)\), and you want to know how far you need to travel to get to the river using the shortest path.
What is the distance from the point to the line? Your answer should be accurate to within \(\pm 0.01\).
Solution
The shortest path will be perpendicular to the line (perpendicular to the river).
Since the river’s slope is \(2\), the path’s slope will be \(\frac{-1}{2}\), because perpendicular lines have slopes that are negative reciprocals. In other words, if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then \(m_1\cdot m_2=-1\).
The point-slope form is useful for generating an equation of a line with a specific slope through a specific point. You can think of it as a vertical stretch and shifts of the \(y=x\) function. Anyway, a line through point \((h,k)\) with slope \(m\) can be represented by \(y=m\cdot(x-h)+k\).
Let function \(r\) represent the river’s line and function \(p\) represent the path’s line. Use point-slope form to represent the path, since we know the path’s slope and a point on the path.
\[r(x)~=~2x-22\]
\[p(x)~=~\frac{-1}{2}(x+11)+16\]
Now, we want to know where the river and path intersect. In other words, we want to find \(x_{\mathrm{i}}\) such that \(r(x_{\mathrm{i}})~=~p(x_{\mathrm{i}})\), where the subscript of \(i\) is used to refer to the intersection.
Multiply both sides by 2 (and distribute to all terms).
\[{4}x_{\mathrm{i}}-44~=~-x_{\mathrm{i}}-11+32\]
Combine the two constants on the right side.
\[ {4}x_{\mathrm{i}}-44~=~-x_{\mathrm{i}}+21 \]
Subtract \(-x_{\mathrm{i}}\) from both sides.
\[ {5}x_{\mathrm{i}}-44~=~21 \]
Subtract \(-44\) from both sides.
\[ {5}x_{\mathrm{i}}~=~65 \]
Divide both sides by \(5\).
\[ x_{\mathrm{i}}~=~13 \]
We need to find \(y_{\mathrm{i}}\), the vertical coordinate of the intersection. You can find evaluating either \(p(13)\) or \(r(13)\).
\[y_{\mathrm{i}}~=~r(13)~=~p(13)\]
I think it is easier to use function \(r\), which was defined above as \(r(x)=2x-22\). We will evaluate \(r(13)\) to get \(y_{\mathrm{i}}\).
\[y_{\mathrm{i}}~=~ r(13)\]
\[y_{\mathrm{i}}~=~ 2(13)-22\]
\[y_{\mathrm{i}}~=~ 26-22\]
\[y_{\mathrm{i}}~=~ 4\]
So, the river and the path intersect at point \((13,4)\). I’ll point out that you can double check this in Desmos by graphing \(y=r(x)\) and \(y=p(x)\).
Now we can use the distance formula to get our answer. Imagine a right triangle with vertices at the two points and with horizontal and vertical legs. We can find absolute differences of corresponding coordinates to find the legs’ lengths.
So, we can use Pythagorean Theorem to find the distance.
\[D^2~=~ 24^2+12^2\]
\[D~=~ \sqrt{24^2+12^2}\]
\[D~=~ \sqrt{576+144}\]
\[D~=~ \sqrt{720}\]
\[D~\approx~ 26.8328157\]
Question
Find the nearest distance from point \((-14,-15)\) to line \(y=-3x+13\).
For flavor, imagine on a map, a river can be approximated by the line \(y=-3x+13\). You are currently at position \((-14,-15)\), and you want to know how far you need to travel to get to the river using the shortest path.
What is the distance from the point to the line? Your answer should be accurate to within \(\pm 0.01\).
Solution
The shortest path will be perpendicular to the line (perpendicular to the river).
Since the river’s slope is \(-3\), the path’s slope will be \(\frac{1}{3}\), because perpendicular lines have slopes that are negative reciprocals. In other words, if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then \(m_1\cdot m_2=-1\).
The point-slope form is useful for generating an equation of a line with a specific slope through a specific point. You can think of it as a vertical stretch and shifts of the \(y=x\) function. Anyway, a line through point \((h,k)\) with slope \(m\) can be represented by \(y=m\cdot(x-h)+k\).
Let function \(r\) represent the river’s line and function \(p\) represent the path’s line. Use point-slope form to represent the path, since we know the path’s slope and a point on the path.
\[r(x)~=~-3x+13\]
\[p(x)~=~\frac{1}{3}(x+14)-15\]
Now, we want to know where the river and path intersect. In other words, we want to find \(x_{\mathrm{i}}\) such that \(r(x_{\mathrm{i}})~=~p(x_{\mathrm{i}})\), where the subscript of \(i\) is used to refer to the intersection.
Multiply both sides by 3 (and distribute to all terms).
\[{-9}x_{\mathrm{i}}+39~=~x_{\mathrm{i}}+14-45\]
Combine the two constants on the right side.
\[ {-9}x_{\mathrm{i}}+39~=~x_{\mathrm{i}}-31 \]
Subtract \(x_{\mathrm{i}}\) from both sides.
\[ {-10}x_{\mathrm{i}}+39~=~-31 \]
Subtract \(39\) from both sides.
\[ {-10}x_{\mathrm{i}}~=~-70 \]
Divide both sides by \(-10\).
\[ x_{\mathrm{i}}~=~7 \]
We need to find \(y_{\mathrm{i}}\), the vertical coordinate of the intersection. You can find evaluating either \(p(7)\) or \(r(7)\).
\[y_{\mathrm{i}}~=~r(7)~=~p(7)\]
I think it is easier to use function \(r\), which was defined above as \(r(x)=-3x+13\). We will evaluate \(r(7)\) to get \(y_{\mathrm{i}}\).
\[y_{\mathrm{i}}~=~ r(7)\]
\[y_{\mathrm{i}}~=~ -3(7)+13\]
\[y_{\mathrm{i}}~=~ -21+13\]
\[y_{\mathrm{i}}~=~ -8\]
So, the river and the path intersect at point \((7,-8)\). I’ll point out that you can double check this in Desmos by graphing \(y=r(x)\) and \(y=p(x)\).
Now we can use the distance formula to get our answer. Imagine a right triangle with vertices at the two points and with horizontal and vertical legs. We can find absolute differences of corresponding coordinates to find the legs’ lengths.
So, we can use Pythagorean Theorem to find the distance.
\[D^2~=~ 21^2+7^2\]
\[D~=~ \sqrt{21^2+7^2}\]
\[D~=~ \sqrt{441+49}\]
\[D~=~ \sqrt{490}\]
\[D~\approx~ 22.1359436\]
Question
Find the nearest distance from point \((16,-4)\) to line \(y=2x-1\).
For flavor, imagine on a map, a river can be approximated by the line \(y=2x-1\). You are currently at position \((16,-4)\), and you want to know how far you need to travel to get to the river using the shortest path.
What is the distance from the point to the line? Your answer should be accurate to within \(\pm 0.01\).
Solution
The shortest path will be perpendicular to the line (perpendicular to the river).
Since the river’s slope is \(2\), the path’s slope will be \(\frac{-1}{2}\), because perpendicular lines have slopes that are negative reciprocals. In other words, if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then \(m_1\cdot m_2=-1\).
The point-slope form is useful for generating an equation of a line with a specific slope through a specific point. You can think of it as a vertical stretch and shifts of the \(y=x\) function. Anyway, a line through point \((h,k)\) with slope \(m\) can be represented by \(y=m\cdot(x-h)+k\).
Let function \(r\) represent the river’s line and function \(p\) represent the path’s line. Use point-slope form to represent the path, since we know the path’s slope and a point on the path.
\[r(x)~=~2x-1\]
\[p(x)~=~\frac{-1}{2}(x-16)-4\]
Now, we want to know where the river and path intersect. In other words, we want to find \(x_{\mathrm{i}}\) such that \(r(x_{\mathrm{i}})~=~p(x_{\mathrm{i}})\), where the subscript of \(i\) is used to refer to the intersection.
Multiply both sides by 2 (and distribute to all terms).
\[{4}x_{\mathrm{i}}-2~=~-x_{\mathrm{i}}+16-8\]
Combine the two constants on the right side.
\[ {4}x_{\mathrm{i}}-2~=~-x_{\mathrm{i}}+8 \]
Subtract \(-x_{\mathrm{i}}\) from both sides.
\[ {5}x_{\mathrm{i}}-2~=~8 \]
Subtract \(-2\) from both sides.
\[ {5}x_{\mathrm{i}}~=~10 \]
Divide both sides by \(5\).
\[ x_{\mathrm{i}}~=~2 \]
We need to find \(y_{\mathrm{i}}\), the vertical coordinate of the intersection. You can find evaluating either \(p(2)\) or \(r(2)\).
\[y_{\mathrm{i}}~=~r(2)~=~p(2)\]
I think it is easier to use function \(r\), which was defined above as \(r(x)=2x-1\). We will evaluate \(r(2)\) to get \(y_{\mathrm{i}}\).
\[y_{\mathrm{i}}~=~ r(2)\]
\[y_{\mathrm{i}}~=~ 2(2)-1\]
\[y_{\mathrm{i}}~=~ 4-1\]
\[y_{\mathrm{i}}~=~ 3\]
So, the river and the path intersect at point \((2,3)\). I’ll point out that you can double check this in Desmos by graphing \(y=r(x)\) and \(y=p(x)\).
Now we can use the distance formula to get our answer. Imagine a right triangle with vertices at the two points and with horizontal and vertical legs. We can find absolute differences of corresponding coordinates to find the legs’ lengths.
So, we can use Pythagorean Theorem to find the distance.
\[D^2~=~ 14^2+7^2\]
\[D~=~ \sqrt{14^2+7^2}\]
\[D~=~ \sqrt{196+49}\]
\[D~=~ \sqrt{245}\]
\[D~\approx~ 15.6524758\]
Question
Find the nearest distance from point \((-17,16)\) to line \(y=\frac{-2}{3}x+22\).
For flavor, imagine on a map, a river can be approximated by the line \(y=\frac{-2}{3}x+22\). You are currently at position \((-17,16)\), and you want to know how far you need to travel to get to the river using the shortest path.
What is the distance from the point to the line? Your answer should be accurate to within \(\pm 0.01\).
Solution
The shortest path will be perpendicular to the line (perpendicular to the river).
Since the river’s slope is \(\frac{-2}{3}\), the path’s slope will be \(\frac{3}{2}\), because perpendicular lines have slopes that are negative reciprocals. In other words, if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then \(m_1\cdot m_2=-1\).
The point-slope form is useful for generating an equation of a line with a specific slope through a specific point. You can think of it as a vertical stretch and shifts of the \(y=x\) function. Anyway, a line through point \((h,k)\) with slope \(m\) can be represented by \(y=m\cdot(x-h)+k\).
Let function \(r\) represent the river’s line and function \(p\) represent the path’s line. Use point-slope form to represent the path, since we know the path’s slope and a point on the path.
\[r(x)~=~\frac{-2}{3}x+22\]
\[p(x)~=~\frac{3}{2}(x+17)+16\]
Now, we want to know where the river and path intersect. In other words, we want to find \(x_{\mathrm{i}}\) such that \(r(x_{\mathrm{i}})~=~p(x_{\mathrm{i}})\), where the subscript of \(i\) is used to refer to the intersection.
Combine the two constants on the right side.
\[ {-4}x_{\mathrm{i}}+132~=~{9}x_{\mathrm{i}}+249 \]
Subtract \({9}x_{\mathrm{i}}\) from both sides.
\[ {-13}x_{\mathrm{i}}+132~=~249 \]
Subtract \(132\) from both sides.
\[ {-13}x_{\mathrm{i}}~=~117 \]
Divide both sides by \(-13\).
\[ x_{\mathrm{i}}~=~-9 \]
We need to find \(y_{\mathrm{i}}\), the vertical coordinate of the intersection. You can find evaluating either \(p(-9)\) or \(r(-9)\).
\[y_{\mathrm{i}}~=~r(-9)~=~p(-9)\]
I think it is easier to use function \(r\), which was defined above as \(r(x)=\frac{-2}{3}x+22\). We will evaluate \(r(-9)\) to get \(y_{\mathrm{i}}\).
\[y_{\mathrm{i}}~=~ r(-9)\]
\[y_{\mathrm{i}}~=~ \frac{-2}{3}(-9)+22\]
\[y_{\mathrm{i}}~=~ 6+22\]
\[y_{\mathrm{i}}~=~ 28\]
So, the river and the path intersect at point \((-9,28)\). I’ll point out that you can double check this in Desmos by graphing \(y=r(x)\) and \(y=p(x)\).
Now we can use the distance formula to get our answer. Imagine a right triangle with vertices at the two points and with horizontal and vertical legs. We can find absolute differences of corresponding coordinates to find the legs’ lengths.
So, we can use Pythagorean Theorem to find the distance.
\[D^2~=~ 8^2+12^2\]
\[D~=~ \sqrt{8^2+12^2}\]
\[D~=~ \sqrt{64+144}\]
\[D~=~ \sqrt{208}\]
\[D~\approx~ 14.4222051\]
Question
Find the nearest distance from point \((-9,-23)\) to line \(y=-2x+4\).
For flavor, imagine on a map, a river can be approximated by the line \(y=-2x+4\). You are currently at position \((-9,-23)\), and you want to know how far you need to travel to get to the river using the shortest path.
What is the distance from the point to the line? Your answer should be accurate to within \(\pm 0.01\).
Solution
The shortest path will be perpendicular to the line (perpendicular to the river).
Since the river’s slope is \(-2\), the path’s slope will be \(\frac{1}{2}\), because perpendicular lines have slopes that are negative reciprocals. In other words, if \(m_1\) and \(m_2\) are slopes of perpendicular lines, then \(m_1\cdot m_2=-1\).
The point-slope form is useful for generating an equation of a line with a specific slope through a specific point. You can think of it as a vertical stretch and shifts of the \(y=x\) function. Anyway, a line through point \((h,k)\) with slope \(m\) can be represented by \(y=m\cdot(x-h)+k\).
Let function \(r\) represent the river’s line and function \(p\) represent the path’s line. Use point-slope form to represent the path, since we know the path’s slope and a point on the path.
\[r(x)~=~-2x+4\]
\[p(x)~=~\frac{1}{2}(x+9)-23\]
Now, we want to know where the river and path intersect. In other words, we want to find \(x_{\mathrm{i}}\) such that \(r(x_{\mathrm{i}})~=~p(x_{\mathrm{i}})\), where the subscript of \(i\) is used to refer to the intersection.
Multiply both sides by 2 (and distribute to all terms).
\[{-4}x_{\mathrm{i}}+8~=~x_{\mathrm{i}}+9-46\]
Combine the two constants on the right side.
\[ {-4}x_{\mathrm{i}}+8~=~x_{\mathrm{i}}-37 \]
Subtract \(x_{\mathrm{i}}\) from both sides.
\[ {-5}x_{\mathrm{i}}+8~=~-37 \]
Subtract \(8\) from both sides.
\[ {-5}x_{\mathrm{i}}~=~-45 \]
Divide both sides by \(-5\).
\[ x_{\mathrm{i}}~=~9 \]
We need to find \(y_{\mathrm{i}}\), the vertical coordinate of the intersection. You can find evaluating either \(p(9)\) or \(r(9)\).
\[y_{\mathrm{i}}~=~r(9)~=~p(9)\]
I think it is easier to use function \(r\), which was defined above as \(r(x)=-2x+4\). We will evaluate \(r(9)\) to get \(y_{\mathrm{i}}\).
\[y_{\mathrm{i}}~=~ r(9)\]
\[y_{\mathrm{i}}~=~ -2(9)+4\]
\[y_{\mathrm{i}}~=~ -18+4\]
\[y_{\mathrm{i}}~=~ -14\]
So, the river and the path intersect at point \((9,-14)\). I’ll point out that you can double check this in Desmos by graphing \(y=r(x)\) and \(y=p(x)\).
Now we can use the distance formula to get our answer. Imagine a right triangle with vertices at the two points and with horizontal and vertical legs. We can find absolute differences of corresponding coordinates to find the legs’ lengths.
So, we can use Pythagorean Theorem to find the distance.
\[D^2~=~ 18^2+9^2\]
\[D~=~ \sqrt{18^2+9^2}\]
\[D~=~ \sqrt{324+81}\]
\[D~=~ \sqrt{405}\]
\[D~\approx~ 20.1246118\]
Question
An event has two types of tickets: each adult ticket costs $5.83 and each child ticket costs $2.11. The event sold 48 tickets for a revenue of $153.36.
How many child tickets were sold?
Solution
Let’s use \(x\) to represent the number of adult tickets and \(y\) to represent the number of child tickets. We know the total number of tickets sold is 48.
\[x+y~=~48\]
The revenue from adults equals the product of $5.83 and \(x\). The revenue from children equals the product of $2.11 and \(y\). The total revenue is the sum of the two types of revenue, and it equals $153.36.
\[5.83x+2.11y~=~153.36\]
We have a system of two equations and two unknowns. We want to solve for \(y\), which is the number of child tickets. To do this, I’d start by isolating \(x\) in the first equation (the equation from the number of tickets).
\[x~=~48-y\]
We can now substitute \(48-y\) for \(x\) in the second equation (the equation from the total revenue).
\[5.83\cdot(48-y)+2.11y=153.36\]
Distribute.
\[279.84-5.83y+2.11y=153.36\]
Subtract 279.84 from both sides.
\[-5.83y+2.11y=-126.48\]
Combine the linear terms.
\[-3.72y=-126.48\]
Divide both sides by -3.72.
\[y=34\]
Question
An event has two types of tickets: each adult ticket costs $5.52 and each child ticket costs $2.14. The event sold 39 tickets for a revenue of $144.30.
How many child tickets were sold?
Solution
Let’s use \(x\) to represent the number of adult tickets and \(y\) to represent the number of child tickets. We know the total number of tickets sold is 39.
\[x+y~=~39\]
The revenue from adults equals the product of $5.52 and \(x\). The revenue from children equals the product of $2.14 and \(y\). The total revenue is the sum of the two types of revenue, and it equals $144.30.
\[5.52x+2.14y~=~144.30\]
We have a system of two equations and two unknowns. We want to solve for \(y\), which is the number of child tickets. To do this, I’d start by isolating \(x\) in the first equation (the equation from the number of tickets).
\[x~=~39-y\]
We can now substitute \(39-y\) for \(x\) in the second equation (the equation from the total revenue).
\[5.52\cdot(39-y)+2.14y=144.30\]
Distribute.
\[215.28-5.52y+2.14y=144.30\]
Subtract 215.28 from both sides.
\[-5.52y+2.14y=-70.98\]
Combine the linear terms.
\[-3.38y=-70.98\]
Divide both sides by -3.38.
\[y=21\]
Question
An event has two types of tickets: each adult ticket costs $7.29 and each child ticket costs $3.10. The event sold 63 tickets for a revenue of $274.91.
How many child tickets were sold?
Solution
Let’s use \(x\) to represent the number of adult tickets and \(y\) to represent the number of child tickets. We know the total number of tickets sold is 63.
\[x+y~=~63\]
The revenue from adults equals the product of $7.29 and \(x\). The revenue from children equals the product of $3.10 and \(y\). The total revenue is the sum of the two types of revenue, and it equals $274.91.
\[7.29x+3.10y~=~274.91\]
We have a system of two equations and two unknowns. We want to solve for \(y\), which is the number of child tickets. To do this, I’d start by isolating \(x\) in the first equation (the equation from the number of tickets).
\[x~=~63-y\]
We can now substitute \(63-y\) for \(x\) in the second equation (the equation from the total revenue).
\[7.29\cdot(63-y)+3.10y=274.91\]
Distribute.
\[459.27-7.29y+3.10y=274.91\]
Subtract 459.27 from both sides.
\[-7.29y+3.10y=-184.36\]
Combine the linear terms.
\[-4.19y=-184.36\]
Divide both sides by -4.19.
\[y=44\]
Question
An event has two types of tickets: each adult ticket costs $6.32 and each child ticket costs $3.89. The event sold 48 tickets for a revenue of $213.45.
How many adult tickets were sold?
Solution
Let’s use \(x\) to represent the number of adult tickets and \(y\) to represent the number of child tickets. We know the total number of tickets sold is 48.
\[x+y~=~48\]
The revenue from adults equals the product of $6.32 and \(x\). The revenue from children equals the product of $3.89 and \(y\). The total revenue is the sum of the two types of revenue, and it equals $213.45.
\[6.32x+3.89y~=~213.45\]
We have a system of two equations and two unknowns. We want to solve for \(x\), which is the number of adult tickets. To do this, I’d start by isolating \(y\) in the first equation (the equation from the number of tickets).
\[y~=~48-x\]
We can now substitute \(48-x\) for \(y\) in the second equation (the equation from the total revenue).
\[6.32x+3.89\cdot(48-x)=213.45\]
Distribute.
\[6.32x+186.72-3.89x=213.45\]
Subtract 186.72 from both sides.
\[6.32x-3.89x=26.73\]
Combine the linear terms.
\[2.43x=26.73\]
Divide both sides by 2.43.
\[x=11\]
Question
An event has two types of tickets: each adult ticket costs $6.77 and each child ticket costs $2.91. The event sold 73 tickets for a revenue of $405.43.
How many child tickets were sold?
Solution
Let’s use \(x\) to represent the number of adult tickets and \(y\) to represent the number of child tickets. We know the total number of tickets sold is 73.
\[x+y~=~73\]
The revenue from adults equals the product of $6.77 and \(x\). The revenue from children equals the product of $2.91 and \(y\). The total revenue is the sum of the two types of revenue, and it equals $405.43.
\[6.77x+2.91y~=~405.43\]
We have a system of two equations and two unknowns. We want to solve for \(y\), which is the number of child tickets. To do this, I’d start by isolating \(x\) in the first equation (the equation from the number of tickets).
\[x~=~73-y\]
We can now substitute \(73-y\) for \(x\) in the second equation (the equation from the total revenue).
\[6.77\cdot(73-y)+2.91y=405.43\]
Distribute.
\[494.21-6.77y+2.91y=405.43\]
Subtract 494.21 from both sides.
\[-6.77y+2.91y=-88.78\]
Combine the linear terms.
\[-3.86y=-88.78\]
Divide both sides by -3.86.
\[y=23\]
Question
The inner dimensions of a frame are 39 inches by 16 inches. An artist hopes to print an image with an aspect ratio of 3.53 such that a uniformly wide mat fits around the image.
Find \(b\), the width of the mat in inches. The tolerance is \(\pm 0.01\) inches.
Solution
A system of equations can be written from the frame’s width and height:
\[39 = 3.53y+2b\]\[16 = y+2b\]
Subtract the two equations.
\[23 = 2.53y\]
Divide both sides by 2.53.
\[9.0909091 = y\]
To find \(b\), you can rearrange the second equation from the system, and plug in the value of \(y\).
\[b ~=~ \frac{16-y}{2} ~=~ \frac{16-9.0909091}{2} ~=~ 3.4545455\]
Question
The inner dimensions of a frame are 31 inches by 20 inches. An artist hopes to print an image with an aspect ratio of 2.33 such that a uniformly wide mat fits around the image.
Find \(b\), the width of the mat in inches. The tolerance is \(\pm 0.01\) inches.
Solution
A system of equations can be written from the frame’s width and height:
\[31 = 2.33y+2b\]\[20 = y+2b\]
Subtract the two equations.
\[11 = 1.33y\]
Divide both sides by 1.33.
\[8.2706767 = y\]
To find \(b\), you can rearrange the second equation from the system, and plug in the value of \(y\).
\[b ~=~ \frac{20-y}{2} ~=~ \frac{20-8.2706767}{2} ~=~ 5.8646617\]
Question
The inner dimensions of a frame are 39 inches by 30 inches. An artist hopes to print an image with an aspect ratio of 1.7 such that a uniformly wide mat fits around the image.
Find \(y\), the height of the image in inches. The tolerance is \(\pm 0.01\) inches.
Solution
A system of equations can be written from the frame’s width and height:
\[39 = 1.7y+2b\]\[30 = y+2b\]
Subtract the two equations.
\[9 = 0.7y\]
Divide both sides by 0.7.
\[12.8571429 = y\]
To find \(b\), you can rearrange the second equation from the system, and plug in the value of \(y\).
\[b ~=~ \frac{30-y}{2} ~=~ \frac{30-12.8571429}{2} ~=~ 8.5714286\]
Question
The inner dimensions of a frame are 29 inches by 23 inches. An artist hopes to print an image with an aspect ratio of 1.71 such that a uniformly wide mat fits around the image.
Find \(b\), the width of the mat in inches. The tolerance is \(\pm 0.01\) inches.
Solution
A system of equations can be written from the frame’s width and height:
\[29 = 1.71y+2b\]\[23 = y+2b\]
Subtract the two equations.
\[6 = 0.71y\]
Divide both sides by 0.71.
\[8.4507042 = y\]
To find \(b\), you can rearrange the second equation from the system, and plug in the value of \(y\).
\[b ~=~ \frac{23-y}{2} ~=~ \frac{23-8.4507042}{2} ~=~ 7.2746479\]
Question
The inner dimensions of a frame are 29 inches by 14 inches. An artist hopes to print an image with an aspect ratio of 4.07 such that a uniformly wide mat fits around the image.
Find \(y\), the height of the image in inches. The tolerance is \(\pm 0.01\) inches.
Solution
A system of equations can be written from the frame’s width and height:
\[29 = 4.07y+2b\]\[14 = y+2b\]
Subtract the two equations.
\[15 = 3.07y\]
Divide both sides by 3.07.
\[4.8859935 = y\]
To find \(b\), you can rearrange the second equation from the system, and plug in the value of \(y\).
\[b ~=~ \frac{14-y}{2} ~=~ \frac{14-4.8859935}{2} ~=~ 4.5570033\]
Question
A lifeguard wants to save a struggling swimmer as soon as possible. The lifeguard can run along the beach at 4 m/s and swim at 1 m/s. The lifeguard will run \(x\) meters and swim \(\sqrt{(210-x)^2+60^2}\) meters.
Find the value of \(x\) (in meters) that minimizes the amount of time. The tolerance is \(\pm 0.1\) meters.
Solution
Use a graphing utility to find the minimum.
So the optimal value of \(x\) equals 194.51 in order to get to the struggling swimmer as quickly as possible.
Question
A lifeguard wants to save a struggling swimmer as soon as possible. The lifeguard can run along the beach at 2.2 m/s and swim at 1.3 m/s. The lifeguard will run \(x\) meters and swim \(\sqrt{(200-x)^2+30^2}\) meters.
Find the value of \(x\) (in meters) that minimizes the amount of time. The tolerance is \(\pm 0.1\) meters.
Solution
Use a graphing utility to find the minimum.
So the optimal value of \(x\) equals 178.03 in order to get to the struggling swimmer as quickly as possible.
Question
A lifeguard wants to save a struggling swimmer as soon as possible. The lifeguard can run along the beach at 3.6 m/s and swim at 0.8 m/s. The lifeguard will run \(x\) meters and swim \(\sqrt{(260-x)^2+40^2}\) meters.
Find the value of \(x\) (in meters) that minimizes the amount of time. The tolerance is \(\pm 0.1\) meters.
Solution
Use a graphing utility to find the minimum.
So the optimal value of \(x\) equals 250.88 in order to get to the struggling swimmer as quickly as possible.
Question
A lifeguard wants to save a struggling swimmer as soon as possible. The lifeguard can run along the beach at 2.5 m/s and swim at 1.4 m/s. The lifeguard will run \(x\) meters and swim \(\sqrt{(210-x)^2+30^2}\) meters.
Find the value of \(x\) (in meters) that minimizes the amount of time. The tolerance is \(\pm 0.1\) meters.
Solution
Use a graphing utility to find the minimum.
So the optimal value of \(x\) equals 189.72 in order to get to the struggling swimmer as quickly as possible.
Question
A lifeguard wants to save a struggling swimmer as soon as possible. The lifeguard can run along the beach at 3.3 m/s and swim at 1.8 m/s. The lifeguard will run \(x\) meters and swim \(\sqrt{(270-x)^2+30^2}\) meters.
Find the value of \(x\) (in meters) that minimizes the amount of time. The tolerance is \(\pm 0.1\) meters.
Solution
Use a graphing utility to find the minimum.
So the optimal value of \(x\) equals 250.48 in order to get to the struggling swimmer as quickly as possible.
Question
A student has taken 5 exams and gotten an average of 70.6. What score does the student need on the next exam to bring the average to 75? (All exams are equally weighted in the average.)
Solution
Find the current total.
\[5 \cdot 70.6 = 353\]
Find the necessary total.
\[6\cdot75 = 450\]
Find the difference of the totals.
\[450-353 = 97\]
The answer is 97.
Question
A student has taken 4 exams and gotten an average of 62.5. What score does the student need on the next exam to bring the average to 70? (All exams are equally weighted in the average.)
Solution
Find the current total.
\[4 \cdot 62.5 = 250\]
Find the necessary total.
\[5\cdot70 = 350\]
Find the difference of the totals.
\[350-250 = 100\]
The answer is 100.
Question
A student has taken 5 exams and gotten an average of 64.6. What score does the student need on the next exam to bring the average to 70? (All exams are equally weighted in the average.)
Solution
Find the current total.
\[5 \cdot 64.6 = 323\]
Find the necessary total.
\[6\cdot70 = 420\]
Find the difference of the totals.
\[420-323 = 97\]
The answer is 97.
Question
A student has taken 3 exams and gotten an average of 70. What score does the student need on the next exam to bring the average to 75? (All exams are equally weighted in the average.)
Solution
Find the current total.
\[3 \cdot 70 = 210\]
Find the necessary total.
\[4\cdot75 = 300\]
Find the difference of the totals.
\[300-210 = 90\]
The answer is 90.
Question
A student has taken 5 exams and gotten an average of 61.2. What score does the student need on the next exam to bring the average to 65? (All exams are equally weighted in the average.)
Solution
Find the current total.
\[5 \cdot 61.2 = 306\]
Find the necessary total.
\[6\cdot65 = 390\]
Find the difference of the totals.
\[390-306 = 84\]
The answer is 84.
Question
A circle is the collection of points equally distant from a center. If the circle has a center \((h,k)\) and a radius \(r\), then we can represent that circle in the \(xy\)-plane using the following standard form of a circle:
\[(x-h)^2+(y-k)^2 = r^2\]
Notice that this equation looks like the Pythagorean Equation (from the Pythagorean Theorem): \(a^2+b^2=c^2\). It also looks like the distance formula, were \(r\) is the distance, \(x-h\) is the horizontal displacement, and \(y-k\) is the vertical displacement.
Consider the circle graphed below. Some of the points on the circle are \((12,-9)\), \((10,-7)\), \((8,-9)\), and \((10,-11)\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The center of the circle is at \((10, -9)\), so \(h=10\) and \(k=-9\). The radius of the circle is the distance from the center to the edge. The radius is 2, so \(r=2\).
The standard equation of the circle is shown:
\[(x-10)^2+(y+9)^2=2^2\]
Question
A circle is the collection of points equally distant from a center. If the circle has a center \((h,k)\) and a radius \(r\), then we can represent that circle in the \(xy\)-plane using the following standard form of a circle:
\[(x-h)^2+(y-k)^2 = r^2\]
Notice that this equation looks like the Pythagorean Equation (from the Pythagorean Theorem): \(a^2+b^2=c^2\). It also looks like the distance formula, were \(r\) is the distance, \(x-h\) is the horizontal displacement, and \(y-k\) is the vertical displacement.
Consider the circle graphed below. Some of the points on the circle are \((11,-2)\), \((4,5)\), \((-3,-2)\), and \((4,-9)\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The center of the circle is at \((4, -2)\), so \(h=4\) and \(k=-2\). The radius of the circle is the distance from the center to the edge. The radius is 7, so \(r=7\).
The standard equation of the circle is shown:
\[(x-4)^2+(y+2)^2=7^2\]
Question
A circle is the collection of points equally distant from a center. If the circle has a center \((h,k)\) and a radius \(r\), then we can represent that circle in the \(xy\)-plane using the following standard form of a circle:
\[(x-h)^2+(y-k)^2 = r^2\]
Notice that this equation looks like the Pythagorean Equation (from the Pythagorean Theorem): \(a^2+b^2=c^2\). It also looks like the distance formula, were \(r\) is the distance, \(x-h\) is the horizontal displacement, and \(y-k\) is the vertical displacement.
Consider the circle graphed below. Some of the points on the circle are \((5,-2)\), \((-4,7)\), \((-13,-2)\), and \((-4,-11)\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The center of the circle is at \((-4, -2)\), so \(h=-4\) and \(k=-2\). The radius of the circle is the distance from the center to the edge. The radius is 9, so \(r=9\).
The standard equation of the circle is shown:
\[(x+4)^2+(y+2)^2=9^2\]
Question
A circle is the collection of points equally distant from a center. If the circle has a center \((h,k)\) and a radius \(r\), then we can represent that circle in the \(xy\)-plane using the following standard form of a circle:
\[(x-h)^2+(y-k)^2 = r^2\]
Notice that this equation looks like the Pythagorean Equation (from the Pythagorean Theorem): \(a^2+b^2=c^2\). It also looks like the distance formula, were \(r\) is the distance, \(x-h\) is the horizontal displacement, and \(y-k\) is the vertical displacement.
Consider the circle graphed below. Some of the points on the circle are \((14,-7)\), \((8,-1)\), \((2,-7)\), and \((8,-13)\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The center of the circle is at \((8, -7)\), so \(h=8\) and \(k=-7\). The radius of the circle is the distance from the center to the edge. The radius is 6, so \(r=6\).
The standard equation of the circle is shown:
\[(x-8)^2+(y+7)^2=6^2\]
Question
A circle is the collection of points equally distant from a center. If the circle has a center \((h,k)\) and a radius \(r\), then we can represent that circle in the \(xy\)-plane using the following standard form of a circle:
\[(x-h)^2+(y-k)^2 = r^2\]
Notice that this equation looks like the Pythagorean Equation (from the Pythagorean Theorem): \(a^2+b^2=c^2\). It also looks like the distance formula, were \(r\) is the distance, \(x-h\) is the horizontal displacement, and \(y-k\) is the vertical displacement.
Consider the circle graphed below. Some of the points on the circle are \((-4,3)\), \((-9,8)\), \((-14,3)\), and \((-9,-2)\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The center of the circle is at \((-9, 3)\), so \(h=-9\) and \(k=3\). The radius of the circle is the distance from the center to the edge. The radius is 5, so \(r=5\).
The standard equation of the circle is shown:
\[(x+9)^2+(y-3)^2=5^2\]
Question
The following equation produces a circle.
\[(x-6)^2+(y+9)^2=121\]
That circle has a center point, \((h,k)\), and radius, \(r\).
Find the values of the parameters \(h\), \(k\), and \(r\).
\(h=\)
\(k=\)
\(r=\)
Solution
The standard form of a circle at center \((h,k)\) with radius \(r\) is:
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.8,0)\) and \((4.8,0)\) and a covertex at \((0, 5.5)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (5.78, 3.36):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.8)^2 + (5.5)^2 =a^2\]
\[a ~=~ \sqrt{4.8^2+5.5^2} ~=~ 7.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,5.5) to \(F_2\)\[2a = 14.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.8\]\[2c = 9.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(5.78-(-4.8))^2+3.36^2}\]
\[s=11.1007207\]\[s\approx11.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(5.78-(4.8))^2+3.36^2}\]
\[w=3.5\]\[w\approx3.5\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=14.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-7.7,0)\) and \((7.7,0)\) and a covertex at \((0, 3.6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (3.86, 3.21):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(7.7)^2 + (3.6)^2 =a^2\]
\[a ~=~ \sqrt{7.7^2+3.6^2} ~=~ 8.5\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,3.6) to \(F_2\)\[2a = 17\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=7.7\]\[2c = 15.4\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(3.86-(-7.7))^2+3.21^2}\]
\[s=11.9974039\]\[s\approx12\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(3.86-(7.7))^2+3.21^2}\]
\[w=5.0049675\]\[w\approx5\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=17\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-7.2,0)\) and \((7.2,0)\) and a covertex at \((0, 6.5)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (7, 4.5):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(7.2)^2 + (6.5)^2 =a^2\]
\[a ~=~ \sqrt{7.2^2+6.5^2} ~=~ 9.7\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,6.5) to \(F_2\)\[2a = 19.4\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=7.2\]\[2c = 14.4\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(7-(-7.2))^2+4.5^2}\]
\[s=14.8959726\]\[s\approx14.9\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(7-(7.2))^2+4.5^2}\]
\[w=4.5044423\]\[w\approx4.5\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=19.4\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.4,0)\) and \((2.4,0)\) and a covertex at \((0, 7)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (5.24, 4.94):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.4)^2 + (7)^2 =a^2\]
\[a ~=~ \sqrt{2.4^2+7^2} ~=~ 7.4\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7) to \(F_2\)\[2a = 14.8\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.4\]\[2c = 4.8\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(5.24-(-2.4))^2+4.94^2}\]
\[s=9.0979778\]\[s\approx9.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(5.24-(2.4))^2+4.94^2}\]
\[w=5.6981751\]\[w\approx5.7\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=14.8\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.5,0)\) and \((4.5,0)\) and a covertex at \((0, 2.8)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (2.47, 2.48):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.5)^2 + (2.8)^2 =a^2\]
\[a ~=~ \sqrt{4.5^2+2.8^2} ~=~ 5.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,2.8) to \(F_2\)\[2a = 10.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.5\]\[2c = 9\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(2.47-(-4.5))^2+2.48^2}\]
\[s=7.3980606\]\[s\approx7.4\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(2.47-(4.5))^2+2.48^2}\]
\[w=3.2048869\]\[w\approx3.2\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=10.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[36x^2+216x+4y^2+32y+244=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[36(x^2+6x)+4(y^2+8y)=-244\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(6\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (9). Notice \((x+3)^2\equiv x^2+6x+9\). Lastly, since we are adding 9 inside parentheses with a multiplier of 36, we are technically adding 324 to both sides to make an equivalent equation.
\[36(x^2+6x+9)+4(y^2+8y)=-244+324\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(8\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (16). Notice \((y+4)^2\equiv y^2+8x+16\). Lastly, since we are adding 16 inside parentheses with a multiplier of 4, we are technically adding 64 to both sides to make an equivalent equation.
\[36(x^2+6x+9)+4(y^2+8y+16)=-244+324+64\]
Combine the constants.
\[36(x^2+6x+9)+4(y^2+8y+16)=144\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[36(x+3)^2+4(y+4)^2=144\]
Divide both sides by \(144\).
\[\frac{36(x+3)^2}{144}+\frac{4(y+4)^2}{144}=1\]
Simplify the fractions. Notice that \(\frac{36}{144}= \frac{1}{4}\) and \(\frac{4}{144}= \frac{1}{36}\).
\[\frac{(x+3)^2}{4}+\frac{(y+4)^2}{36}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{4}=2\) and \(\sqrt{36}=6\).
\[\frac{(x+3)^2}{2^2}+\frac{(y+4)^2}{6^2}=1\]
Thus,
\[\begin{align}
h &= -3 \\
k &= -4 \\
r_1 &= 2 \\
r_2 &= 6
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[25x^2+150x+4y^2+8y+129=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[25x^2+150x+4y^2+8y+129=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[25(x^2+6x)+4(y^2+2y)=-129\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(6\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (9). Notice \((x+3)^2\equiv x^2+6x+9\). Lastly, since we are adding 9 inside parentheses with a multiplier of 25, we are technically adding 225 to both sides to make an equivalent equation.
\[25(x^2+6x+9)+4(y^2+2y)=-129+225\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y+1)^2\equiv y^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 4, we are technically adding 4 to both sides to make an equivalent equation.
\[25(x^2+6x+9)+4(y^2+2y+1)=-129+225+4\]
Combine the constants.
\[25(x^2+6x+9)+4(y^2+2y+1)=100\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[25(x+3)^2+4(y+1)^2=100\]
Divide both sides by \(100\).
\[\frac{25(x+3)^2}{100}+\frac{4(y+1)^2}{100}=1\]
Simplify the fractions. Notice that \(\frac{25}{100}= \frac{1}{4}\) and \(\frac{4}{100}= \frac{1}{25}\).
\[\frac{(x+3)^2}{4}+\frac{(y+1)^2}{25}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{4}=2\) and \(\sqrt{25}=5\).
\[\frac{(x+3)^2}{2^2}+\frac{(y+1)^2}{5^2}=1\]
Thus,
\[\begin{align}
h &= -3 \\
k &= -1 \\
r_1 &= 2 \\
r_2 &= 5
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[64x^2+768x+4y^2-8y+2052=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[64x^2+768x+4y^2-8y+2052=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[64(x^2+12x)+4(y^2-2y)=-2052\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(12\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (36). Notice \((x+6)^2\equiv x^2+12x+36\). Lastly, since we are adding 36 inside parentheses with a multiplier of 64, we are technically adding 2304 to both sides to make an equivalent equation.
\[64(x^2+12x+36)+4(y^2-2y)=-2052+2304\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y-1)^2\equiv y^2-2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 4, we are technically adding 4 to both sides to make an equivalent equation.
\[64(x^2+12x+36)+4(y^2-2y+1)=-2052+2304+4\]
Combine the constants.
\[64(x^2+12x+36)+4(y^2-2y+1)=256\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[64(x+6)^2+4(y-1)^2=256\]
Divide both sides by \(256\).
\[\frac{64(x+6)^2}{256}+\frac{4(y-1)^2}{256}=1\]
Simplify the fractions. Notice that \(\frac{64}{256}= \frac{1}{4}\) and \(\frac{4}{256}= \frac{1}{64}\).
\[\frac{(x+6)^2}{4}+\frac{(y-1)^2}{64}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{4}=2\) and \(\sqrt{64}=8\).
\[\frac{(x+6)^2}{2^2}+\frac{(y-1)^2}{8^2}=1\]
Thus,
\[\begin{align}
h &= -6 \\
k &= 1 \\
r_1 &= 2 \\
r_2 &= 8
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[9x^2+90x+4y^2+48y+333=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[9x^2+90x+4y^2+48y+333=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[9(x^2+10x)+4(y^2+12y)=-333\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(10\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (25). Notice \((x+5)^2\equiv x^2+10x+25\). Lastly, since we are adding 25 inside parentheses with a multiplier of 9, we are technically adding 225 to both sides to make an equivalent equation.
\[9(x^2+10x+25)+4(y^2+12y)=-333+225\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(12\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (36). Notice \((y+6)^2\equiv y^2+12x+36\). Lastly, since we are adding 36 inside parentheses with a multiplier of 4, we are technically adding 144 to both sides to make an equivalent equation.
\[9(x^2+10x+25)+4(y^2+12y+36)=-333+225+144\]
Combine the constants.
\[9(x^2+10x+25)+4(y^2+12y+36)=36\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[9(x+5)^2+4(y+6)^2=36\]
Divide both sides by \(36\).
\[\frac{9(x+5)^2}{36}+\frac{4(y+6)^2}{36}=1\]
Simplify the fractions. Notice that \(\frac{9}{36}= \frac{1}{4}\) and \(\frac{4}{36}= \frac{1}{9}\).
\[\frac{(x+5)^2}{4}+\frac{(y+6)^2}{9}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{4}=2\) and \(\sqrt{9}=3\).
\[\frac{(x+5)^2}{2^2}+\frac{(y+6)^2}{3^2}=1\]
Thus,
\[\begin{align}
h &= -5 \\
k &= -6 \\
r_1 &= 2 \\
r_2 &= 3
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[16x^2+32x+49y^2-490y+457=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[16x^2+32x+49y^2-490y+457=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[16(x^2+2x)+49(y^2-10y)=-457\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x+1)^2\equiv x^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 16, we are technically adding 16 to both sides to make an equivalent equation.
\[16(x^2+2x+1)+49(y^2-10y)=-457+16\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-10\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (25). Notice \((y-5)^2\equiv y^2-10x+25\). Lastly, since we are adding 25 inside parentheses with a multiplier of 49, we are technically adding 1225 to both sides to make an equivalent equation.
\[16(x^2+2x+1)+49(y^2-10y+25)=-457+16+1225\]
Combine the constants.
\[16(x^2+2x+1)+49(y^2-10y+25)=784\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[16(x+1)^2+49(y-5)^2=784\]
Divide both sides by \(784\).
\[\frac{16(x+1)^2}{784}+\frac{49(y-5)^2}{784}=1\]
Simplify the fractions. Notice that \(\frac{16}{784}= \frac{1}{49}\) and \(\frac{49}{784}= \frac{1}{16}\).
\[\frac{(x+1)^2}{49}+\frac{(y-5)^2}{16}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{49}=7\) and \(\sqrt{16}=4\).
\[\frac{(x+1)^2}{7^2}+\frac{(y-5)^2}{4^2}=1\]
Thus,
\[\begin{align}
h &= -1 \\
k &= 5 \\
r_1 &= 7 \\
r_2 &= 4
\end{align}\]
Question
An alien creature flies a round trip — from home, to school, and back home — using the exact same route both directions (so the distances of the initial and return trips are equal).
On the initial trip, the creature maintained a speed of 84 meters per second. On the return trip, the creature maintained a speed of 51 meters per second.
What was the average speed, in m/s, of the entire trip? (The tolerance is \(\pm 0.01\) m/s.)
Solution
You might incorrectly guess the average speed for the trip is simply \(\frac{84+51}{2}\), but you’d be wrong.
The “easy” way to do this is to pick a distance. You can later check that you’ll get the same answer for any distance.
I’ll pick a distance of 1000 meters for each trip.
Time of initial trip = \(\frac{1000}{84}=11.9047619\) seconds.
Time of return trip = \(\frac{1000}{51}=19.6078431\) seconds.
The total distance traveled is 2000 meters. The total time is \(11.9047619+19.6078431=31.512605\) seconds. To find the average speed, divide the total distance by the total time.
Of course, there is a cooler way to do this. Let’s use the following variables:
\[\begin{align}
s_1 &= \text{speed to school}\\
s_2 &= \text{speed returning home}\\
D &= \text{distance}\\
t_1 &= \text{time to get to school}\\
t_2 &= \text{time to return home}\\
\bar{s} &= \text{average speed for round trip}
\end{align}\]
If you combine all three of the above equations, you should be able to make a formula for \(\bar{s}\) that only depends on \(s_1\) and \(s_2\).
Question
An alien creature flies a round trip — from home, to school, and back home — using the exact same route both directions (so the distances of the initial and return trips are equal).
On the initial trip, the creature maintained a speed of 24 meters per second. On the return trip, the creature maintained a speed of 62 meters per second.
What was the average speed, in m/s, of the entire trip? (The tolerance is \(\pm 0.01\) m/s.)
Solution
You might incorrectly guess the average speed for the trip is simply \(\frac{24+62}{2}\), but you’d be wrong.
The “easy” way to do this is to pick a distance. You can later check that you’ll get the same answer for any distance.
I’ll pick a distance of 1000 meters for each trip.
Time of initial trip = \(\frac{1000}{24}=41.6666667\) seconds.
Time of return trip = \(\frac{1000}{62}=16.1290323\) seconds.
The total distance traveled is 2000 meters. The total time is \(41.6666667+16.1290323=57.7956989\) seconds. To find the average speed, divide the total distance by the total time.
Of course, there is a cooler way to do this. Let’s use the following variables:
\[\begin{align}
s_1 &= \text{speed to school}\\
s_2 &= \text{speed returning home}\\
D &= \text{distance}\\
t_1 &= \text{time to get to school}\\
t_2 &= \text{time to return home}\\
\bar{s} &= \text{average speed for round trip}
\end{align}\]
If you combine all three of the above equations, you should be able to make a formula for \(\bar{s}\) that only depends on \(s_1\) and \(s_2\).
Question
An alien creature flies a round trip — from home, to school, and back home — using the exact same route both directions (so the distances of the initial and return trips are equal).
On the initial trip, the creature maintained a speed of 65 meters per second. On the return trip, the creature maintained a speed of 80 meters per second.
What was the average speed, in m/s, of the entire trip? (The tolerance is \(\pm 0.01\) m/s.)
Solution
You might incorrectly guess the average speed for the trip is simply \(\frac{65+80}{2}\), but you’d be wrong.
The “easy” way to do this is to pick a distance. You can later check that you’ll get the same answer for any distance.
I’ll pick a distance of 1000 meters for each trip.
Time of initial trip = \(\frac{1000}{65}=15.3846154\) seconds.
Time of return trip = \(\frac{1000}{80}=12.5\) seconds.
The total distance traveled is 2000 meters. The total time is \(15.3846154+12.5=27.8846154\) seconds. To find the average speed, divide the total distance by the total time.
Of course, there is a cooler way to do this. Let’s use the following variables:
\[\begin{align}
s_1 &= \text{speed to school}\\
s_2 &= \text{speed returning home}\\
D &= \text{distance}\\
t_1 &= \text{time to get to school}\\
t_2 &= \text{time to return home}\\
\bar{s} &= \text{average speed for round trip}
\end{align}\]
If you combine all three of the above equations, you should be able to make a formula for \(\bar{s}\) that only depends on \(s_1\) and \(s_2\).
Question
An alien creature flies a round trip — from home, to school, and back home — using the exact same route both directions (so the distances of the initial and return trips are equal).
On the initial trip, the creature maintained a speed of 88 meters per second. On the return trip, the creature maintained a speed of 73 meters per second.
What was the average speed, in m/s, of the entire trip? (The tolerance is \(\pm 0.01\) m/s.)
Solution
You might incorrectly guess the average speed for the trip is simply \(\frac{88+73}{2}\), but you’d be wrong.
The “easy” way to do this is to pick a distance. You can later check that you’ll get the same answer for any distance.
I’ll pick a distance of 1000 meters for each trip.
Time of initial trip = \(\frac{1000}{88}=11.3636364\) seconds.
Time of return trip = \(\frac{1000}{73}=13.6986301\) seconds.
The total distance traveled is 2000 meters. The total time is \(11.3636364+13.6986301=25.0622665\) seconds. To find the average speed, divide the total distance by the total time.
Of course, there is a cooler way to do this. Let’s use the following variables:
\[\begin{align}
s_1 &= \text{speed to school}\\
s_2 &= \text{speed returning home}\\
D &= \text{distance}\\
t_1 &= \text{time to get to school}\\
t_2 &= \text{time to return home}\\
\bar{s} &= \text{average speed for round trip}
\end{align}\]
If you combine all three of the above equations, you should be able to make a formula for \(\bar{s}\) that only depends on \(s_1\) and \(s_2\).
Question
An alien creature flies a round trip — from home, to school, and back home — using the exact same route both directions (so the distances of the initial and return trips are equal).
On the initial trip, the creature maintained a speed of 39 meters per second. On the return trip, the creature maintained a speed of 35 meters per second.
What was the average speed, in m/s, of the entire trip? (The tolerance is \(\pm 0.01\) m/s.)
Solution
You might incorrectly guess the average speed for the trip is simply \(\frac{39+35}{2}\), but you’d be wrong.
The “easy” way to do this is to pick a distance. You can later check that you’ll get the same answer for any distance.
I’ll pick a distance of 1000 meters for each trip.
Time of initial trip = \(\frac{1000}{39}=25.6410256\) seconds.
Time of return trip = \(\frac{1000}{35}=28.5714286\) seconds.
The total distance traveled is 2000 meters. The total time is \(25.6410256+28.5714286=54.2124542\) seconds. To find the average speed, divide the total distance by the total time.
Of course, there is a cooler way to do this. Let’s use the following variables:
\[\begin{align}
s_1 &= \text{speed to school}\\
s_2 &= \text{speed returning home}\\
D &= \text{distance}\\
t_1 &= \text{time to get to school}\\
t_2 &= \text{time to return home}\\
\bar{s} &= \text{average speed for round trip}
\end{align}\]
You can probably guess correctly how this would generalize with more points, but this seems tedious enough. Eh, what the heck… if there were four points, it would look like:
Plug those values into the Lagrange polynomial for three points. I would type this function (and the three points) into Desmos. Make sure the curve hits all three points!
You can probably guess correctly how this would generalize with more points, but this seems tedious enough. Eh, what the heck… if there were four points, it would look like:
Plug those values into the Lagrange polynomial for three points. I would type this function (and the three points) into Desmos. Make sure the curve hits all three points!
You can probably guess correctly how this would generalize with more points, but this seems tedious enough. Eh, what the heck… if there were four points, it would look like:
Plug those values into the Lagrange polynomial for three points. I would type this function (and the three points) into Desmos. Make sure the curve hits all three points!
You can probably guess correctly how this would generalize with more points, but this seems tedious enough. Eh, what the heck… if there were four points, it would look like:
Plug those values into the Lagrange polynomial for three points. I would type this function (and the three points) into Desmos. Make sure the curve hits all three points!
You can probably guess correctly how this would generalize with more points, but this seems tedious enough. Eh, what the heck… if there were four points, it would look like:
Plug those values into the Lagrange polynomial for three points. I would type this function (and the three points) into Desmos. Make sure the curve hits all three points!
In Desmos, we can get best-fit curves by using their least-squares regression tool. This type of analysis is very important in science and research.
In this problem, I want you to run a cubic regression on the following points:
x
y
3
7
5
6
6
7
7
8
8
4
The result of the regression will be a cubic polynomial, which we can call \(f\). You will then evaluate \(f(9)\).
If you copy/paste the above table into a Desmos item, it should create a table. If not, add a new table and add the points shown above.
In that item’s top-left corner, there should be a “Add Regression” button; click it. Change the regression type to “Cubic Regression”. You’ll see a cubic curve that does its best to get as close to the points as possible. You can “Export a snapshot to the expression list” by clicking on the button next to EQUATION. In that new expression, replace the \(y\) with \(f(x)\).
Now, in a new item, you should be able to evaluate \(f(9)\). What do you get? (We are using regression to make a prediction. The tolerance is \(\pm0.01\).)
Solution
Your cubic regression should give the following polynomial.
In Desmos, we can get best-fit curves by using their least-squares regression tool. This type of analysis is very important in science and research.
In this problem, I want you to run a cubic regression on the following points:
x
y
3
1
5
2
7
3
8
3
9
6
The result of the regression will be a cubic polynomial, which we can call \(f\). You will then evaluate \(f(1)\).
If you copy/paste the above table into a Desmos item, it should create a table. If not, add a new table and add the points shown above.
In that item’s top-left corner, there should be a “Add Regression” button; click it. Change the regression type to “Cubic Regression”. You’ll see a cubic curve that does its best to get as close to the points as possible. You can “Export a snapshot to the expression list” by clicking on the button next to EQUATION. In that new expression, replace the \(y\) with \(f(x)\).
Now, in a new item, you should be able to evaluate \(f(1)\). What do you get? (We are using regression to make a prediction. The tolerance is \(\pm0.01\).)
Solution
Your cubic regression should give the following polynomial.
In Desmos, we can get best-fit curves by using their least-squares regression tool. This type of analysis is very important in science and research.
In this problem, I want you to run a cubic regression on the following points:
x
y
1
6
2
1
3
5
4
8
7
9
The result of the regression will be a cubic polynomial, which we can call \(f\). You will then evaluate \(f(6)\).
If you copy/paste the above table into a Desmos item, it should create a table. If not, add a new table and add the points shown above.
In that item’s top-left corner, there should be a “Add Regression” button; click it. Change the regression type to “Cubic Regression”. You’ll see a cubic curve that does its best to get as close to the points as possible. You can “Export a snapshot to the expression list” by clicking on the button next to EQUATION. In that new expression, replace the \(y\) with \(f(x)\).
Now, in a new item, you should be able to evaluate \(f(6)\). What do you get? (We are using regression to make a prediction. The tolerance is \(\pm0.01\).)
Solution
Your cubic regression should give the following polynomial.
In Desmos, we can get best-fit curves by using their least-squares regression tool. This type of analysis is very important in science and research.
In this problem, I want you to run a cubic regression on the following points:
x
y
2
8
3
1
5
2
6
2
8
5
The result of the regression will be a cubic polynomial, which we can call \(f\). You will then evaluate \(f(1)\).
If you copy/paste the above table into a Desmos item, it should create a table. If not, add a new table and add the points shown above.
In that item’s top-left corner, there should be a “Add Regression” button; click it. Change the regression type to “Cubic Regression”. You’ll see a cubic curve that does its best to get as close to the points as possible. You can “Export a snapshot to the expression list” by clicking on the button next to EQUATION. In that new expression, replace the \(y\) with \(f(x)\).
Now, in a new item, you should be able to evaluate \(f(1)\). What do you get? (We are using regression to make a prediction. The tolerance is \(\pm0.01\).)
Solution
Your cubic regression should give the following polynomial.
In Desmos, we can get best-fit curves by using their least-squares regression tool. This type of analysis is very important in science and research.
In this problem, I want you to run a cubic regression on the following points:
x
y
2
3
3
4
5
5
6
4
8
3
The result of the regression will be a cubic polynomial, which we can call \(f\). You will then evaluate \(f(1)\).
If you copy/paste the above table into a Desmos item, it should create a table. If not, add a new table and add the points shown above.
In that item’s top-left corner, there should be a “Add Regression” button; click it. Change the regression type to “Cubic Regression”. You’ll see a cubic curve that does its best to get as close to the points as possible. You can “Export a snapshot to the expression list” by clicking on the button next to EQUATION. In that new expression, replace the \(y\) with \(f(x)\).
Now, in a new item, you should be able to evaluate \(f(1)\). What do you get? (We are using regression to make a prediction. The tolerance is \(\pm0.01\).)
Solution
Your cubic regression should give the following polynomial.
Using the linear model, predict \(y\) when \(x=41\).
Solution
Copy the \(x\) and \(y\) values into a spreadsheet.
Highlight the data. Insert chart \(\to\) scatterplot.
Under Setup, click Use column A as labels
Under Customize \(\to\) Series, click Trendline
Change the Label to Use Equation
The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.
A
B
C
1
x
y
2
39.2
283.5
=linest(B2:B8, A2:A8, TRUE, FALSE)
3
39.1
291.8
4
31.8
234.5
5
38
268.1
6
38.4
286.8
7
36.2
253.7
8
36.4
249.1
That linest will provide two numbers: the slope 7.6425428 and the intercept -16.0975474.
So, our linear model is:
\[y = (7.6425428)x + (-16.0975474) \]
Plug in the given value of \(x=41\) for prediction.
Using the linear model, predict \(y\) when \(x=75\).
Solution
Copy the \(x\) and \(y\) values into a spreadsheet.
Highlight the data. Insert chart \(\to\) scatterplot.
Under Setup, click Use column A as labels
Under Customize \(\to\) Series, click Trendline
Change the Label to Use Equation
The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.
A
B
C
1
x
y
2
65.1
282.3
=linest(B2:B10, A2:A10, TRUE, FALSE)
3
61.8
279
4
63.2
281.5
5
60
268.1
6
57.7
255.9
7
62.6
272
8
63.9
274.7
9
59.1
262.3
10
66.5
292.9
That linest will provide two numbers: the slope 3.6757952 and the intercept 45.6246944.
So, our linear model is:
\[y = (3.6757952)x + (45.6246944) \]
Plug in the given value of \(x=75\) for prediction.
Using the linear model, predict \(y\) when \(x=32\).
Solution
Copy the \(x\) and \(y\) values into a spreadsheet.
Highlight the data. Insert chart \(\to\) scatterplot.
Under Setup, click Use column A as labels
Under Customize \(\to\) Series, click Trendline
Change the Label to Use Equation
The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.
A
B
C
1
x
y
2
41.9
63.9
=linest(B2:B10, A2:A10, TRUE, FALSE)
3
46
67.3
4
49.7
78
5
42.9
61.5
6
46.4
70.4
7
43.5
62.6
8
43.5
68.8
9
48.2
78.6
10
50.1
83.2
That linest will provide two numbers: the slope 2.3834729 and the intercept -38.6852792.
So, our linear model is:
\[y = (2.3834729)x + (-38.6852792) \]
Plug in the given value of \(x=32\) for prediction.
Using the linear model, predict \(y\) when \(x=48\).
Solution
Copy the \(x\) and \(y\) values into a spreadsheet.
Highlight the data. Insert chart \(\to\) scatterplot.
Under Setup, click Use column A as labels
Under Customize \(\to\) Series, click Trendline
Change the Label to Use Equation
The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.
A
B
C
1
x
y
2
50.7
422.9
=linest(B2:B10, A2:A10, TRUE, FALSE)
3
60.3
505.4
4
48.9
403
5
57.4
457.6
6
57.4
484.1
7
60.9
520.3
8
53.3
448.9
9
48.9
406.4
10
59.9
498.8
That linest will provide two numbers: the slope 8.7345156 and the intercept -22.1964884.
So, our linear model is:
\[y = (8.7345156)x + (-22.1964884) \]
Plug in the given value of \(x=48\) for prediction.
Using the linear model, predict \(y\) when \(x=97\).
Solution
Copy the \(x\) and \(y\) values into a spreadsheet.
Highlight the data. Insert chart \(\to\) scatterplot.
Under Setup, click Use column A as labels
Under Customize \(\to\) Series, click Trendline
Change the Label to Use Equation
The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.
A
B
C
1
x
y
2
87.1
279
=linest(B2:B7, A2:A7, TRUE, FALSE)
3
89
284
4
87.4
279
5
85.7
277.9
6
88.7
288.7
7
88.6
288.9
That linest will provide two numbers: the slope 3.2936358 and the intercept -6.099874.
So, our linear model is:
\[y = (3.2936358)x + (-6.099874) \]
Plug in the given value of \(x=97\) for prediction.
